barycentric coordinates of the foot of an altitude in a tetrahedron Given three sides $a=BC$, $b=CA$, $c=AB$ of a triangle $ABC$. The foot $H$ on the segment $BC$ of the altitude $AH$ can be computed via (its proof is easy, using the Law of Cosine, for example)
$$H=\dfrac{1}{a^2+c^2-b^2}B+\dfrac{1}{a^2+b^2-c^2}C.\tag{1}$$
In other words, the barycentric coordinate of $H$ with respect to $BC$ is
$$\left(\dfrac{1}{a^2+c^2-b^2}:\dfrac{1}{a^2+b^2-c^2}\right).\tag{2}$$
Note that there is no $A$ in the above formulas.
As a consequence, we have another way of construction a triangle when 3 sides is given: take $C=(0,0)$, $B=(a,0)$, then take $H$ using that barycentric formula, and $A$ can be obtained by the Pythagorean theorem.
Now construction a tetrahedron knowing $6$ sides leads to the following question.

Question: Knowing three sides $a=BC$, $b=CA$, $c=AB$ of the base $ABC$` and three edges $a'=DA$, $b'=DB$, $c'=DC$, how can we calculate the foot $H$ on the base $ABC$ of the altitude $DH$ without using apex $D$ ?
I guess there is natural formula for 3-dimensional analog. However, I have not found the answer for that question. Numerical checking show that the formula
$$H=\dfrac{1}{S_b^2+S_c^2-S_a^2}A+\dfrac{1}{S_c^2+S_a^2-S_b^2}B+\dfrac{1}{S_a^2+S_b^2-S_c^2}C$$
does not hold, where $S_b=S_{\Delta DCA}$, $S_c=S_{\Delta DAB}$, $S_a=S_{\Delta DBC}$ are areas of lateral faces (that can be calculated via $a$, $b$, $c$, $a'$, $b'$, $c'$).
Thanks for any helps!
 A: CAVEAT. This answer is based on the first version of the question, where $=$, $=$ and $=$. Probably the OP didn't intend to set $t$ as a variable, but as a sort of subscript, and so they changed to $'=$, $'=$ and $'=$. Anyway, I'm not going to revise my answer: it contains a detailed explanation, which should enable interested readers to make any change they like.

Draw altitudes $DE$ and $DF$ in faces $ACD$ and $BCD$. If we set $a_1=CF$ and $b_1=CE$, from Pythagorean theorem we get:
$$
a_1={t^2c^2-t^2b^2+a^2\over 2a},
\quad
b_1={t^2c^2-t^2a^2+b^2\over 2b}.
$$
In quadrilateral $CEHF$, if we set $\gamma=\angle BCA$ and $\phi=\angle BCH$,
from $CH=a_1/\cos\phi=b_1/\cos(\gamma-\phi)$ we can find $\tan\phi$ and then obtain:
$$
FH={b_1-a_1\cos\gamma\over\sin\gamma},
\quad
EH={a_1-b_1\cos\gamma\over\sin\gamma},
\quad
CH={\sqrt{a_1^2+b_1^2-2a_1b_1\cos\gamma}\over\sin\gamma}.
$$

EDIT.
Using the above results one can compute the barycentric coordinates $(x_A,x_B,x_C)$ for point $H=x_AA+x_BB+x_CC$.
For instance, with the aid of Mathematica, one gets for $x_A$:
$$
x_A=
\frac{a^4 - a^2 b^2 - a^2 c^2 + 
(2 a^4 - a^2 b^2 - b^4 - a^2 c^2 + 2 b^2 c^2 - c^4) t^2}
{a^4-2 a^2 b^2-2 a^2 c^2+b^4-2 b^2 c^2+c^4},
$$
while $x_B$ and $x_C$ can be obtained from $x_A$ with the obvious permutations of symbols.
